91Ó°ÊÓ

The Laplace transform is an essential mathematical tool for solving differential equations, particularly those that appear in engineering and physics. It transforms a function of time, usually denoted as \(f(t)\), into a function of a complex variable \(s\), denoted as \(F(s)\). This transformation simplifies many operations, such as derivatives and integrals, by converting them into algebraic equivalents.

The basic formula for the Laplace transform is: \[ F(s) = \int_{0}^{\infty} e^{-st} f(t) dt \].

When applied to the given second-order differential equation, our goal is to turn the differential operation into a simple algebraic form. By taking the Laplace transform of both sides, we turn the problem of differentiation into multiplication by \(s\), making it easier to solve for \(X_a(s)\), the Laplace transform of \(x_a(t)\).

For this exercise, we assume zero initial conditions: \( x_a(0) = 0 \) and \( \frac{dx_a}{dt} (0) = 0 \). These assumptions simplify the transformed equation considerably, allowing us to isolate \(X_a(s)\).

A step function, also known as the Heaviside function, is a piecewise function that jumps from one value to another at a specified point. In this exercise, the input function \(x_e(t)\) is defined as:
\[ x_e(t) = \begin{cases} 0 & \text{für } t \leq 0 \ 1 & \text{für } t > 0 \end{cases} \]

This means that \(x_e(t)\) remains zero until \t\ equals 0 and then instantly jumps to 1 for all times \(t > 0\). The Laplace transform of the step function is particularly straightforward:
\[ L\{x_e(t)\} = \frac{1}{s} \]

This function is critical in control systems and signal processing because it represents a sudden change or pulse. By using the Laplace transform on this step function, we can easily incorporate it into our differential equation to find a solution.

Initial conditions specify the state of a system at the beginning of observation, usually at \(t=0\). These are crucial for solving differential equations because they allow us to determine the specific solution that fits the problem.

In this particular exercise, the initial conditions are:

• \( \lim_{t \rightarrow 0+} x_a(t) = 0 \)
• \( \lim_{t \rightarrow 0+} \frac{dx_a}{dt} = 0 \)

These conditions simplify the problem when applying the Laplace transform, as terms involving initial values of \(x_a(t)\) and its derivatives become zero. As a result, they lead us to the simplified algebraic expression we can solve more easily.

Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions that are easier to work with. This method is particularly helpful when taking the inverse Laplace transform.

After applying the Laplace transform and isolating \(X_a(s)\), we obtain:
\[ X_a(s) = \frac{K}{s(T_1 s^2 + T_2 s + 1)} \]

To find \(x_a(t)\), we need to decompose this fraction into simpler parts that are easier to inverse-transform. We do this by solving for the roots of the quadratic equation \(T_1 s^2 + T_2 s + 1 = 0\) and then expressing our function based on these roots:
\[ X_a(s) = \frac{A}{s} + \frac{B}{s - \lambda_1} + \frac{C}{s - \lambda_2} \]

Here, \( \lambda_1 \) and \( \lambda_2 \) are the solutions to the quadratic equation. By determining the coefficients \(A, B,\text{ and } C \), we break down \(X_a(s)\) into simpler terms, making it easier to find the inverse Laplace transform and, consequently, the time-domain solution \(x_a(t)\).